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Exercise 12.2 · Q14

Q.Prove p→(q→r)≡(p∧q)→rp\to(q\to r)\equiv (p\wedge q)\to r without using truth table.

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We rewrite every conditional using A→B≡¬A∨BA\to B\equiv\neg A\vee B, simplify the resulting disjunction/conjunction using the standard laws (including a De Morgan step), and rewrite the result back as a single conditional.

Step 1. Rewrite the left side's inner conditional. q→r≡¬q∨rq\to r\equiv\neg q\vee r.

Step 2. Rewrite the left side's outer conditional. p→(q→r)≡p→(¬q∨r)≡¬p∨(¬q∨r)p\to(q\to r)\equiv p\to(\neg q\vee r)\equiv\neg p\vee(\neg q\vee r).

Step 3. Regroup using the Associative Law. ¬p∨(¬q∨r)≡(¬p∨¬q)∨r\neg p\vee(\neg q\vee r)\equiv(\neg p\vee\neg q)\vee r.

Step 4. Rewrite the right side. (p∧q)→r≡¬(p∧q)∨r(p\wedge q)\to r\equiv\neg(p\wedge q)\vee r. …

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